Suppose we have a data set with five predictors, X1 = GPA, X2 = IQ, X3 = Gender (1 for Female and 0 for Male), X4 = Interaction between GPA and IQ, and X5 = Interaction between GPA and Gender. The response is starting salary after graduation (in thousands of dollars). Suppose we use least squares to fit the model, and get βˆ0 = 50, βˆ1 = 2 0 , βˆ 2 = 0 . 0 7 , βˆ 3 = 3 5 , βˆ 4 = 0 . 0 1 , βˆ 5 = − 1 0 .

Y = 50 + 20(gpa) + 0.07(iq) + 35(gender) + 0.01(gpa * iq) - 10 (gpa * gender)

  1. For a fixed value of IQ and GPA,males earn more on average than females provided that the GPA is high enough.

  1. Predict the salary of a female with IQ of 110 and a GPA of 4.0. =85+40+7.7+4.4=137.1

  2. True or false: Since the coefficient for the GPA/IQ interaction term is very small, there is very little evidence of an interaction effect. Justify your answer.

False. The p-value of the regression coefficient should be looked at.

3.7.4 (a) X and Y have a linear relationship. Therefore, thier least squares line will be near the true regression line. Further, the RSS of the linear regression will be lower than cubic regression

  1. Test should be used - so (a)

  2. A highly flexible model will closly fit observations and reduce train RSS

  3. Test should be used - so (c)

3.7.5

setwd("~/Dropbox/Public/w4702/Datasets")

3.7.8 a. Use the lm() function to perform a simple linear regression with mpg as the response and horsepower as the predictor. Use the summary() function to print the results. Comment on the output.

library(ISLR)
data(Auto)
fit <- lm(mpg ~ horsepower, data = Auto)
summary(fit)
## 
## Call:
## lm(formula = mpg ~ horsepower, data = Auto)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -13.5710  -3.2592  -0.3435   2.7630  16.9240 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 39.935861   0.717499   55.66   <2e-16 ***
## horsepower  -0.157845   0.006446  -24.49   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.906 on 390 degrees of freedom
## Multiple R-squared:  0.6059, Adjusted R-squared:  0.6049 
## F-statistic: 599.7 on 1 and 390 DF,  p-value: < 2.2e-16
  1. Is there a relationship between the predictor and the re- sponse?

YES. The F statistic is larger than 1 and p-value is close to zero. This is evidence against the null hypothesis and implies there is a relationship between mpg and horsepower.

  1. How strong is the relationship between the predictor and the response?

As the Adjusted R-squared is 0.605, 60.5% of the varinace in mpg is dependent on horsepower

  1. Is the relationship between the predictor and the response positive or negative?

The mpg is inversly proportional to horsepower - i.e. high horsepower leads to lower mpg.

  1. What is the predicted mpg associated with a horsepower of 98? What are the associated 95 % confidence and prediction intervals?
predict(fit, data.frame(horsepower = 98), interval = "confidence")
##        fit      lwr      upr
## 1 24.46708 23.97308 24.96108
predict(fit, data.frame(horsepower = 98), interval = "prediction")
##        fit     lwr      upr
## 1 24.46708 14.8094 34.12476
  1. Plot the response and the predictor. Use the abline() function to display the least squares regression line.
plot(Auto$horsepower, Auto$mpg, main = "Scatterplot comparing mpg and horsepower", xlab = "horsepower", ylab = "mpg", col = "green")
abline(fit, col = "blue")

  1. Use the plot() function to produce diagnostic plots of the least squares regression fit. Comment on any problems you see with the fit.
par(mfrow=c(2,2))
plot(fit)

The residuals sugest a non linear realtionship.

3.7.9 This question involves the use of multiple linear regression on the Auto data set. (a) Produce a scatterplot matrix which includes all of the variables in the data set.

pairs(Auto)

  1. Compute the matrix of correlations between the variables using the function cor(). You will need to exclude the name variable, which is qualitative.
names(Auto)
## [1] "mpg"          "cylinders"    "displacement" "horsepower"  
## [5] "weight"       "acceleration" "year"         "origin"      
## [9] "name"
cor(Auto[1:8])
##                     mpg  cylinders displacement horsepower     weight
## mpg           1.0000000 -0.7776175   -0.8051269 -0.7784268 -0.8322442
## cylinders    -0.7776175  1.0000000    0.9508233  0.8429834  0.8975273
## displacement -0.8051269  0.9508233    1.0000000  0.8972570  0.9329944
## horsepower   -0.7784268  0.8429834    0.8972570  1.0000000  0.8645377
## weight       -0.8322442  0.8975273    0.9329944  0.8645377  1.0000000
## acceleration  0.4233285 -0.5046834   -0.5438005 -0.6891955 -0.4168392
## year          0.5805410 -0.3456474   -0.3698552 -0.4163615 -0.3091199
## origin        0.5652088 -0.5689316   -0.6145351 -0.4551715 -0.5850054
##              acceleration       year     origin
## mpg             0.4233285  0.5805410  0.5652088
## cylinders      -0.5046834 -0.3456474 -0.5689316
## displacement   -0.5438005 -0.3698552 -0.6145351
## horsepower     -0.6891955 -0.4163615 -0.4551715
## weight         -0.4168392 -0.3091199 -0.5850054
## acceleration    1.0000000  0.2903161  0.2127458
## year            0.2903161  1.0000000  0.1815277
## origin          0.2127458  0.1815277  1.0000000
  1. Use the lm() function to perform a multiple linear regression with mpg as the response and all other variables except name as the predictors. Use the summary() function to print the results. Comment on the output. For instance:
  1. Is there a relationship between the predictors and the re- sponse?
lm.fit1 = lm(mpg~.-name, data=Auto)
summary(lm.fit1)
## 
## Call:
## lm(formula = mpg ~ . - name, data = Auto)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -9.5903 -2.1565 -0.1169  1.8690 13.0604 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -17.218435   4.644294  -3.707  0.00024 ***
## cylinders     -0.493376   0.323282  -1.526  0.12780    
## displacement   0.019896   0.007515   2.647  0.00844 ** 
## horsepower    -0.016951   0.013787  -1.230  0.21963    
## weight        -0.006474   0.000652  -9.929  < 2e-16 ***
## acceleration   0.080576   0.098845   0.815  0.41548    
## year           0.750773   0.050973  14.729  < 2e-16 ***
## origin         1.426141   0.278136   5.127 4.67e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 3.328 on 384 degrees of freedom
## Multiple R-squared:  0.8215, Adjusted R-squared:  0.8182 
## F-statistic: 252.4 on 7 and 384 DF,  p-value: < 2.2e-16

The F statistic is larger than 1 and p-value is close to zero. This is evidence against the null hypothesis and implies there is a relationship between mpg and horsepower.

  1. Which predictors appear to have a statistically significant relationship to the response? Comparing the p-values of predictors, we can see that horsepower, accelration and displacement do not have a statistically significant relationship.

  2. What does the coefficient for the year variable suggest? The coefficient for ‘year’ suggests that cars become more fuel efficient over time. Every year the efficiency increases by by about 1mpg.

  1. Use the plot() function to produce diagnostic plots of the linear regression fit. Comment on any problems you see with the fit. Do the residual plots suggest any unusually large outliers? Does the leverage plot identify any observations with unusually high leverage?
par(mfrow=c(2,2))
plot(lm.fit1)

The residuals sugest a non linear realtionship with some outliers.

  1. Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?
fit2 <- lm(mpg ~ cylinders * displacement+displacement * weight, data = Auto[, 1:8])
summary(fit2)
## 
## Call:
## lm(formula = mpg ~ cylinders * displacement + displacement * 
##     weight, data = Auto[, 1:8])
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -13.2934  -2.5184  -0.3476   1.8399  17.7723 
## 
## Coefficients:
##                          Estimate Std. Error t value Pr(>|t|)    
## (Intercept)             5.262e+01  2.237e+00  23.519  < 2e-16 ***
## cylinders               7.606e-01  7.669e-01   0.992    0.322    
## displacement           -7.351e-02  1.669e-02  -4.403 1.38e-05 ***
## weight                 -9.888e-03  1.329e-03  -7.438 6.69e-13 ***
## cylinders:displacement -2.986e-03  3.426e-03  -0.872    0.384    
## displacement:weight     2.128e-05  5.002e-06   4.254 2.64e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.103 on 386 degrees of freedom
## Multiple R-squared:  0.7272, Adjusted R-squared:  0.7237 
## F-statistic: 205.8 on 5 and 386 DF,  p-value: < 2.2e-16

The p-values for displacement and weight show statistically significant relationship while cylinder and displacement is not.

  1. Try a few different transformations of the variables, such as log(X), √X, X2. Comment on your findings. cor()
par(mfrow = c(2, 2))
plot(log(Auto$horsepower), Auto$mpg)
plot(sqrt(Auto$horsepower), Auto$mpg)
plot((Auto$horsepower)^2, Auto$mpg)

Using one of the predictors, we observe non linearity except for log(x). This is consistent with our expectation.

3.7.10 This question should be answered using the Carseats data set. (a) Fit a multiple regression model to predict Sales using Price,Urban, and US.

library(ISLR)
summary(Carseats)
##      Sales          CompPrice       Income        Advertising    
##  Min.   : 0.000   Min.   : 77   Min.   : 21.00   Min.   : 0.000  
##  1st Qu.: 5.390   1st Qu.:115   1st Qu.: 42.75   1st Qu.: 0.000  
##  Median : 7.490   Median :125   Median : 69.00   Median : 5.000  
##  Mean   : 7.496   Mean   :125   Mean   : 68.66   Mean   : 6.635  
##  3rd Qu.: 9.320   3rd Qu.:135   3rd Qu.: 91.00   3rd Qu.:12.000  
##  Max.   :16.270   Max.   :175   Max.   :120.00   Max.   :29.000  
##    Population        Price        ShelveLoc        Age       
##  Min.   : 10.0   Min.   : 24.0   Bad   : 96   Min.   :25.00  
##  1st Qu.:139.0   1st Qu.:100.0   Good  : 85   1st Qu.:39.75  
##  Median :272.0   Median :117.0   Medium:219   Median :54.50  
##  Mean   :264.8   Mean   :115.8                Mean   :53.32  
##  3rd Qu.:398.5   3rd Qu.:131.0                3rd Qu.:66.00  
##  Max.   :509.0   Max.   :191.0                Max.   :80.00  
##    Education    Urban       US     
##  Min.   :10.0   No :118   No :142  
##  1st Qu.:12.0   Yes:282   Yes:258  
##  Median :14.0                      
##  Mean   :13.9                      
##  3rd Qu.:16.0                      
##  Max.   :18.0
  1. Provide an interpretation of each coefficient in the model. Be careful—some of the variables in the model are qualitative!
attach(Carseats)
lm.fit = lm(Sales~Price+Urban+US)
summary(lm.fit)
## 
## Call:
## lm(formula = Sales ~ Price + Urban + US)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -6.9206 -1.6220 -0.0564  1.5786  7.0581 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 13.043469   0.651012  20.036  < 2e-16 ***
## Price       -0.054459   0.005242 -10.389  < 2e-16 ***
## UrbanYes    -0.021916   0.271650  -0.081    0.936    
## USYes        1.200573   0.259042   4.635 4.86e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.472 on 396 degrees of freedom
## Multiple R-squared:  0.2393, Adjusted R-squared:  0.2335 
## F-statistic: 41.52 on 3 and 396 DF,  p-value: < 2.2e-16

Based on the coefficients, on average for every dollar increase in price the sales decrease by 54.45 units An Urban store location and US have a statistically significant relationship to sales.

  1. Write out the model in equation form, being careful to handle the qualitative variables properly.

Sales = 13.04 + -0.05 Price + -0.02 UrbanYes + 1.20 USYes + ε

  1. For which of the predictors can you reject the null hypothesis H0 :βj =0?

The null hypothesis can be rejected for US and Price.

  1. On the basis of your response to the previous question, fit a smaller model that only uses the predictors for which there is evidence of association with the outcome.
lm.fit1 = lm(Sales ~ Price + US)
summary(lm.fit1)
## 
## Call:
## lm(formula = Sales ~ Price + US)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -6.9269 -1.6286 -0.0574  1.5766  7.0515 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 13.03079    0.63098  20.652  < 2e-16 ***
## Price       -0.05448    0.00523 -10.416  < 2e-16 ***
## USYes        1.19964    0.25846   4.641 4.71e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.469 on 397 degrees of freedom
## Multiple R-squared:  0.2393, Adjusted R-squared:  0.2354 
## F-statistic: 62.43 on 2 and 397 DF,  p-value: < 2.2e-16

  1. How well do the models in (a) and (e) fit the data? The linear regression is marginally better.

  2. Using the model from (e), obtain 95% confidence intervals for the coefficient(s).

confint(fit2)
##                                2.5 %        97.5 %
## (Intercept)            48.2243049523  5.702251e+01
## cylinders              -0.7472803670  2.268561e+00
## displacement           -0.1063365850 -4.068896e-02
## weight                 -0.0125019928 -7.274341e-03
## cylinders:displacement -0.0097214570  3.749355e-03
## displacement:weight     0.0000114434  3.111142e-05
  1. Is there evidence of outliers or high leverage observations in the model from (e)?
plot(predict(lm.fit), rstudent(lm.fit))

par(mfrow = c(2, 2))
plot(fit2)

There are some outlierr as reflected by above +2 and below-2 residuals.

3.7.11 In this problem we will investigate the t-statistic for the null hypoth- esis H0 : β = 0 in simple linear regression without an intercept. To begin, we generate a predictor x and a response y as follows.

set.seed(1)
x <- rnorm(100)
y <- 2 * x + rnorm(100)
  1. Perform a simple linear regression of y onto x, without an in- tercept. Report the coefficient estimate βˆ, the standard error of this coefficient estimate, and the t-statistic and p-value associ- ated with the null hypothesis H0 : β = 0. Comment on these results. (You can perform regression without an intercept using the command lm(y∼x+0).)
fit4 <- lm(y ~ x + 0)
summary(fit4)
## 
## Call:
## lm(formula = y ~ x + 0)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.9154 -0.6472 -0.1771  0.5056  2.3109 
## 
## Coefficients:
##   Estimate Std. Error t value Pr(>|t|)    
## x   1.9939     0.1065   18.73   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.9586 on 99 degrees of freedom
## Multiple R-squared:  0.7798, Adjusted R-squared:  0.7776 
## F-statistic: 350.7 on 1 and 99 DF,  p-value: < 2.2e-16

As per the summary, we can reject the null based on the small p-value

  1. Now perform a simple linear regression of x onto y without an intercept, and report the coefficient estimate, its standard error, and the corresponding t-statistic and p-values associated with the null hypothesis H0 : β = 0. Comment on these results.
fit5 <- lm(x ~ y + 0)
summary(fit5)
## 
## Call:
## lm(formula = x ~ y + 0)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -0.8699 -0.2368  0.1030  0.2858  0.8938 
## 
## Coefficients:
##   Estimate Std. Error t value Pr(>|t|)    
## y  0.39111    0.02089   18.73   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4246 on 99 degrees of freedom
## Multiple R-squared:  0.7798, Adjusted R-squared:  0.7776 
## F-statistic: 350.7 on 1 and 99 DF,  p-value: < 2.2e-16

As per the summary, we can reject the null based on the small p-value

  1. What is the relationship between the results obtained in (a) and (b)?

Both (a) and (b) are the same line created or y=2x+ε can also be written x=0.5(y−ε)

  1. For the regression of Y onto X without an intercept, the t- statistic for H0 : β = 0 takes the form βˆ/SE(βˆ), where βˆ is given by (3.38).
(sqrt(length(x)-1) * sum(x*y)) / (sqrt(sum(x*x) * sum(y*y) - (sum(x*y))^2))
## [1] 18.72593
  1. Using the results from (d), argue that the t-statistic for the regression of y onto x is the same t-statistic for the regression of x onto y.

Switch x and y will get us the same results.

  1. In R, show that when regression is performed with an intercept, the t-statistic for H0 : β1 = 0 is the same for the regression of y onto x as it is for the regression of x onto y.
fit6 <- lm(y ~ x)
summary(fit6)
## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -1.8768 -0.6138 -0.1395  0.5394  2.3462 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.03769    0.09699  -0.389    0.698    
## x            1.99894    0.10773  18.556   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.9628 on 98 degrees of freedom
## Multiple R-squared:  0.7784, Adjusted R-squared:  0.7762 
## F-statistic: 344.3 on 1 and 98 DF,  p-value: < 2.2e-16
fit7 <- lm(x ~ y)
summary(fit7)
## 
## Call:
## lm(formula = x ~ y)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.90848 -0.28101  0.06274  0.24570  0.85736 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  0.03880    0.04266    0.91    0.365    
## y            0.38942    0.02099   18.56   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4249 on 98 degrees of freedom
## Multiple R-squared:  0.7784, Adjusted R-squared:  0.7762 
## F-statistic: 344.3 on 1 and 98 DF,  p-value: < 2.2e-16

As we can see, the t-statistic for both is the same

3.7.13 In this exercise you will create some simulated data and will fit simple linear regression models to it. Make sure to use set.seed(1) prior to starting part (a) to ensure conistent results.

  1. Using the rnorm() function, create a vector, x, containing 100 observations drawn from a N(0,1) distribution. This represents a feature, X.
  1. Using the rnorm() function, create a vector, eps, containing 100 observations drawn from a N(0,0.25) distribution i.e. a normal distribution with mean zero and variance 0.25.
  2. Using x and eps, generate a vector y according to the model Y =−1+0.5X+ε. (3.39) What is the length of the vector y? What are the values of β0 and β1 in this linear model?
set.seed(1)
x <- rnorm(100)
eps <- rnorm(100, sd = sqrt(0.25))
y <- -1 + 0.5 * x + eps
length(y)
## [1] 100

y is 100 and β0 is -1, β1 is 0.5.

  1. Create a scatterplot displaying the relationship between x and y. Comment on what you observe.
plot(x, y)

Plot shows a linear relationship.

  1. Fit a least squares linear model to predict y using x. Comment on the model obtained. How do βˆ0 and βˆ1 compare to β0 and β1?
lm.fit8 = lm(y~x)
summary(lm.fit8)
## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.93842 -0.30688 -0.06975  0.26970  1.17309 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -1.01885    0.04849 -21.010  < 2e-16 ***
## x            0.49947    0.05386   9.273 4.58e-15 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4814 on 98 degrees of freedom
## Multiple R-squared:  0.4674, Adjusted R-squared:  0.4619 
## F-statistic: 85.99 on 1 and 98 DF,  p-value: 4.583e-15

The F statistic is larger than 1 and p-value is close to zero. This is evidence against the null hypothesis and implies there is a relationship between mpg and horsepower.

  1. Display the least squares line on the scatterplot obtained in (d). Draw the population regression line on the plot, in a different color. Use the legend() command to create an appropriate leg- end.
plot(x, y)
abline(lm.fit8, col = "green")
abline(-1, 0.5, col = "blue")

  1. Now fit a polynomial regression model that predicts y using x and x2. Is there evidence that the quadratic term improves the model fit?
lm.fit_sq = lm(y~x+I(x^2))
summary(lm.fit_sq)
## 
## Call:
## lm(formula = y ~ x + I(x^2))
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.98252 -0.31270 -0.06441  0.29014  1.13500 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.97164    0.05883 -16.517  < 2e-16 ***
## x            0.50858    0.05399   9.420  2.4e-15 ***
## I(x^2)      -0.05946    0.04238  -1.403    0.164    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.479 on 97 degrees of freedom
## Multiple R-squared:  0.4779, Adjusted R-squared:  0.4672 
## F-statistic:  44.4 on 2 and 97 DF,  p-value: 2.038e-14

Based on the t-statistics and the p-value, we can infer that there is no statisticaly relevant relationship between x^2 and y.

  1. Repeat (a)–(f) after modifying the data generation process in such a way that there is less noise in the data. The model (3.39) should remain the same. You can do this by decreasing the vari- ance of the normal distribution used to generate the error term ε in (b). Describe your results.
set.seed(1)
eps <- rnorm(100, sd = 0.125)
x <- rnorm(100)
y <- -1 + 0.5 * x + eps
plot(x, y)
fit9 <- lm(y ~ x)
summary(fit9)
## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.29052 -0.07545  0.00067  0.07288  0.28664 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.98639    0.01129  -87.34   <2e-16 ***
## x            0.49988    0.01184   42.22   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.1128 on 98 degrees of freedom
## Multiple R-squared:  0.9479, Adjusted R-squared:  0.9474 
## F-statistic:  1782 on 1 and 98 DF,  p-value: < 2.2e-16
abline(fit9, col = "green")
abline(-1, 0.5, col = "blue")

The two line overlap as we have little noise

  1. Repeat (a)–(f) after modifying the data generation process in such a way that there is more noise in the data. The model (3.39) should remain the same. You can do this by increasing the variance of the normal distribution used to generate the error term ε in (b). Describe your results.
set.seed(1)
eps <- rnorm(100, sd = 0.5)
x <- rnorm(100)
y <- -1 + 0.5 * x + eps
plot(x, y)
fit10 <- lm(y ~ x)
summary(fit10)
## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.16208 -0.30181  0.00268  0.29152  1.14658 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.94557    0.04517  -20.93   <2e-16 ***
## x            0.49953    0.04736   10.55   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4514 on 98 degrees of freedom
## Multiple R-squared:  0.5317, Adjusted R-squared:  0.5269 
## F-statistic: 111.2 on 1 and 98 DF,  p-value: < 2.2e-16
abline(fit10, col = "green")
abline(-1, 0.5, col = "blue")

The lines are further apart in this case.

  1. What are the confidence intervals for β0 and β1 based on the original data set, the noisier data set, and the less noisy data set? Comment on your results.
confint(lm.fit8)
##                  2.5 %     97.5 %
## (Intercept) -1.1150804 -0.9226122
## x            0.3925794  0.6063602
confint(fit9)
##                 2.5 %     97.5 %
## (Intercept) -1.008805 -0.9639819
## x            0.476387  0.5233799
confint(fit10)
##                  2.5 %     97.5 %
## (Intercept) -1.0352203 -0.8559276
## x            0.4055479  0.5935197

The interval is centered on 0.5.

3.7.14 This problem focuses on the collinearity problem. (a) Perform the following commands in R:

set.seed(1)
x1 <- runif(100)
x2 <- 0.5 * x1 + rnorm(100)/10
y <- 2 + 2 * x1 + 0.3 * x2 + rnorm(100)

The last line corresponds to creating a linear model in which y is a function of x1 and x2. Write out the form of the linear model. What are the regression coefficients? Y=2+2X1+0.3X2+ε The regression coefficients are 2, 2 and 0.3.

  1. What is the correlation between x1 and x2? Create a scatterplot displaying the relationship between the variables.
cor(x1, x2)
## [1] 0.8351212
plot(x1, x2)

They are tightly corelated

  1. Using this data, fit a least squares regression to predict y using x1 and x2. Describe the results obtained. What are βˆ0, βˆ1, and βˆ2? How do these relate to the true β0, β1, and β2? Can you reject the null hypothesis H0 : β1 = 0? How about the null hypothesis H0 : β2 = 0?
fit11 <- lm(y ~ x1 + x2)
summary(fit11)
## 
## Call:
## lm(formula = y ~ x1 + x2)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.8311 -0.7273 -0.0537  0.6338  2.3359 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.1305     0.2319   9.188 7.61e-15 ***
## x1            1.4396     0.7212   1.996   0.0487 *  
## x2            1.0097     1.1337   0.891   0.3754    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.056 on 97 degrees of freedom
## Multiple R-squared:  0.2088, Adjusted R-squared:  0.1925 
## F-statistic:  12.8 on 2 and 97 DF,  p-value: 1.164e-05
fit12 <- lm(y ~ x1)
summary(fit12)
## 
## Call:
## lm(formula = y ~ x1)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.89495 -0.66874 -0.07785  0.59221  2.45560 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.1124     0.2307   9.155 8.27e-15 ***
## x1            1.9759     0.3963   4.986 2.66e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.055 on 98 degrees of freedom
## Multiple R-squared:  0.2024, Adjusted R-squared:  0.1942 
## F-statistic: 24.86 on 1 and 98 DF,  p-value: 2.661e-06
fit13 <- lm(y ~ x2)
summary(fit13)
## 
## Call:
## lm(formula = y ~ x2)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.62687 -0.75156 -0.03598  0.72383  2.44890 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.3899     0.1949   12.26  < 2e-16 ***
## x2            2.8996     0.6330    4.58 1.37e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.072 on 98 degrees of freedom
## Multiple R-squared:  0.1763, Adjusted R-squared:  0.1679 
## F-statistic: 20.98 on 1 and 98 DF,  p-value: 1.366e-05

  1. No, as they are closly corelated.

x1 <- c(x1, 0.1)
x2 <- c(x2, 0.8)
y <- c(y, 6)
fit14 <- lm(y ~ x1 + x2)
fit15 <- lm(y ~ x1)
fit16 <- lm(y ~ x2)
summary(fit14)
## 
## Call:
## lm(formula = y ~ x1 + x2)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.73348 -0.69318 -0.05263  0.66385  2.30619 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.2267     0.2314   9.624 7.91e-16 ***
## x1            0.5394     0.5922   0.911  0.36458    
## x2            2.5146     0.8977   2.801  0.00614 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.075 on 98 degrees of freedom
## Multiple R-squared:  0.2188, Adjusted R-squared:  0.2029 
## F-statistic: 13.72 on 2 and 98 DF,  p-value: 5.564e-06
summary(fit15)
## 
## Call:
## lm(formula = y ~ x1)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -2.8897 -0.6556 -0.0909  0.5682  3.5665 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.2569     0.2390   9.445 1.78e-15 ***
## x1            1.7657     0.4124   4.282 4.29e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.111 on 99 degrees of freedom
## Multiple R-squared:  0.1562, Adjusted R-squared:  0.1477 
## F-statistic: 18.33 on 1 and 99 DF,  p-value: 4.295e-05
summary(fit16)
## 
## Call:
## lm(formula = y ~ x2)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.64729 -0.71021 -0.06899  0.72699  2.38074 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   2.3451     0.1912  12.264  < 2e-16 ***
## x2            3.1190     0.6040   5.164 1.25e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.074 on 99 degrees of freedom
## Multiple R-squared:  0.2122, Adjusted R-squared:  0.2042 
## F-statistic: 26.66 on 1 and 99 DF,  p-value: 1.253e-06
plot(fit14)

plot(fit15)

plot(fit16)

3.7.15 This problem involves the “Boston” data set, which we saw in the lab for this chapter. We will now try to predict per capita crime rate using the other variables in this data set. In other words, per capita crime rate is the response, and the other variables are the predictors.

For each predictor, fit a simple linear regression model to predict the response. Describe your results. In which of the models is there a statistically significant association between the predictor and the response ? Create some plots to back up your assertions.

library(MASS)
attach(Boston)
fit.zn <- lm(crim ~ zn)
summary(fit.zn)
## 
## Call:
## lm(formula = crim ~ zn)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -4.429 -4.222 -2.620  1.250 84.523 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  4.45369    0.41722  10.675  < 2e-16 ***
## zn          -0.07393    0.01609  -4.594 5.51e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.435 on 504 degrees of freedom
## Multiple R-squared:  0.04019,    Adjusted R-squared:  0.03828 
## F-statistic:  21.1 on 1 and 504 DF,  p-value: 5.506e-06
fit.indus <- lm(crim ~ indus)
summary(fit.indus)
## 
## Call:
## lm(formula = crim ~ indus)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -11.972  -2.698  -0.736   0.712  81.813 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -2.06374    0.66723  -3.093  0.00209 ** 
## indus        0.50978    0.05102   9.991  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.866 on 504 degrees of freedom
## Multiple R-squared:  0.1653, Adjusted R-squared:  0.1637 
## F-statistic: 99.82 on 1 and 504 DF,  p-value: < 2.2e-16
chas <- as.factor(chas)
fit.chas <- lm(crim ~ chas)
summary(fit.chas)
## 
## Call:
## lm(formula = crim ~ chas)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -3.738 -3.661 -3.435  0.018 85.232 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   3.7444     0.3961   9.453   <2e-16 ***
## chas1        -1.8928     1.5061  -1.257    0.209    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.597 on 504 degrees of freedom
## Multiple R-squared:  0.003124,   Adjusted R-squared:  0.001146 
## F-statistic: 1.579 on 1 and 504 DF,  p-value: 0.2094
fit.nox <- lm(crim ~ nox)
summary(fit.nox)
## 
## Call:
## lm(formula = crim ~ nox)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -12.371  -2.738  -0.974   0.559  81.728 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  -13.720      1.699  -8.073 5.08e-15 ***
## nox           31.249      2.999  10.419  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.81 on 504 degrees of freedom
## Multiple R-squared:  0.1772, Adjusted R-squared:  0.1756 
## F-statistic: 108.6 on 1 and 504 DF,  p-value: < 2.2e-16
fit.rm <- lm(crim ~ rm)
summary(fit.rm)
## 
## Call:
## lm(formula = crim ~ rm)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -6.604 -3.952 -2.654  0.989 87.197 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   20.482      3.365   6.088 2.27e-09 ***
## rm            -2.684      0.532  -5.045 6.35e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.401 on 504 degrees of freedom
## Multiple R-squared:  0.04807,    Adjusted R-squared:  0.04618 
## F-statistic: 25.45 on 1 and 504 DF,  p-value: 6.347e-07
fit.age <- lm(crim ~ age)
summary(fit.age)
## 
## Call:
## lm(formula = crim ~ age)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -6.789 -4.257 -1.230  1.527 82.849 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -3.77791    0.94398  -4.002 7.22e-05 ***
## age          0.10779    0.01274   8.463 2.85e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.057 on 504 degrees of freedom
## Multiple R-squared:  0.1244, Adjusted R-squared:  0.1227 
## F-statistic: 71.62 on 1 and 504 DF,  p-value: 2.855e-16
fit.dis <- lm(crim ~ dis)
summary(fit.dis)
## 
## Call:
## lm(formula = crim ~ dis)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -6.708 -4.134 -1.527  1.516 81.674 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   9.4993     0.7304  13.006   <2e-16 ***
## dis          -1.5509     0.1683  -9.213   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.965 on 504 degrees of freedom
## Multiple R-squared:  0.1441, Adjusted R-squared:  0.1425 
## F-statistic: 84.89 on 1 and 504 DF,  p-value: < 2.2e-16
fit.rad <- lm(crim ~ rad)
summary(fit.rad)
## 
## Call:
## lm(formula = crim ~ rad)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -10.164  -1.381  -0.141   0.660  76.433 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -2.28716    0.44348  -5.157 3.61e-07 ***
## rad          0.61791    0.03433  17.998  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.718 on 504 degrees of freedom
## Multiple R-squared:  0.3913, Adjusted R-squared:   0.39 
## F-statistic: 323.9 on 1 and 504 DF,  p-value: < 2.2e-16
fit.tax <- lm(crim ~ tax)
summary(fit.tax)
## 
## Call:
## lm(formula = crim ~ tax)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -12.513  -2.738  -0.194   1.065  77.696 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -8.528369   0.815809  -10.45   <2e-16 ***
## tax          0.029742   0.001847   16.10   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.997 on 504 degrees of freedom
## Multiple R-squared:  0.3396, Adjusted R-squared:  0.3383 
## F-statistic: 259.2 on 1 and 504 DF,  p-value: < 2.2e-16
fit.ptratio <- lm(crim ~ ptratio)
summary(fit.ptratio)
## 
## Call:
## lm(formula = crim ~ ptratio)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -7.654 -3.985 -1.912  1.825 83.353 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -17.6469     3.1473  -5.607 3.40e-08 ***
## ptratio       1.1520     0.1694   6.801 2.94e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.24 on 504 degrees of freedom
## Multiple R-squared:  0.08407,    Adjusted R-squared:  0.08225 
## F-statistic: 46.26 on 1 and 504 DF,  p-value: 2.943e-11
fit.black <- lm(crim ~ black)
summary(fit.black)
## 
## Call:
## lm(formula = crim ~ black)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -13.756  -2.299  -2.095  -1.296  86.822 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 16.553529   1.425903  11.609   <2e-16 ***
## black       -0.036280   0.003873  -9.367   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.946 on 504 degrees of freedom
## Multiple R-squared:  0.1483, Adjusted R-squared:  0.1466 
## F-statistic: 87.74 on 1 and 504 DF,  p-value: < 2.2e-16
fit.lstat <- lm(crim ~ lstat)
summary(fit.lstat)
## 
## Call:
## lm(formula = crim ~ lstat)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -13.925  -2.822  -0.664   1.079  82.862 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -3.33054    0.69376  -4.801 2.09e-06 ***
## lstat        0.54880    0.04776  11.491  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.664 on 504 degrees of freedom
## Multiple R-squared:  0.2076, Adjusted R-squared:  0.206 
## F-statistic:   132 on 1 and 504 DF,  p-value: < 2.2e-16
fit.medv <- lm(crim ~ medv)
summary(fit.medv)
## 
## Call:
## lm(formula = crim ~ medv)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -9.071 -4.022 -2.343  1.298 80.957 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 11.79654    0.93419   12.63   <2e-16 ***
## medv        -0.36316    0.03839   -9.46   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.934 on 504 degrees of freedom
## Multiple R-squared:  0.1508, Adjusted R-squared:  0.1491 
## F-statistic: 89.49 on 1 and 504 DF,  p-value: < 2.2e-16
  1. Fit a multiple regression model to predict the response using all of the predictors. Describe your results. For which predictors can we reject the null hypothesis H0 : βj = 0?
fit.all <- lm(crim ~ ., data = Boston)
summary(fit.all)
## 
## Call:
## lm(formula = crim ~ ., data = Boston)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -9.924 -2.120 -0.353  1.019 75.051 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  17.033228   7.234903   2.354 0.018949 *  
## zn            0.044855   0.018734   2.394 0.017025 *  
## indus        -0.063855   0.083407  -0.766 0.444294    
## chas         -0.749134   1.180147  -0.635 0.525867    
## nox         -10.313535   5.275536  -1.955 0.051152 .  
## rm            0.430131   0.612830   0.702 0.483089    
## age           0.001452   0.017925   0.081 0.935488    
## dis          -0.987176   0.281817  -3.503 0.000502 ***
## rad           0.588209   0.088049   6.680 6.46e-11 ***
## tax          -0.003780   0.005156  -0.733 0.463793    
## ptratio      -0.271081   0.186450  -1.454 0.146611    
## black        -0.007538   0.003673  -2.052 0.040702 *  
## lstat         0.126211   0.075725   1.667 0.096208 .  
## medv         -0.198887   0.060516  -3.287 0.001087 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.439 on 492 degrees of freedom
## Multiple R-squared:  0.454,  Adjusted R-squared:  0.4396 
## F-statistic: 31.47 on 13 and 492 DF,  p-value: < 2.2e-16

zn, dis, rad, black, medv

  1. How do your results from (a) compare to your results from (b)? Create a plot displaying the univariate regression coefficients from (a) on the x-axis, and the multiple regression coefficients from (b) on the y-axis. That is, each predictor is displayed as a single point in the plot. Its coefficient in a simple linear regres- sion model is shown on the x-axis, and its coefficient estimate in the multiple linear regression model is shown on the y-axis.
x = c(coefficients(fit.zn)[2],
      coefficients(fit.indus)[2],
      coefficients(fit.chas)[2],
      coefficients(fit.nox)[2],
      coefficients(fit.rm)[2],
      coefficients(fit.age)[2],
      coefficients(fit.dis)[2],
      coefficients(fit.rad)[2],
      coefficients(fit.tax)[2],
      coefficients(fit.ptratio)[2],
      coefficients(fit.black)[2],
      coefficients(fit.lstat)[2],
      coefficients(fit.medv)[2])
y = coefficients(fit.all)[2:14]
plot(x, y)

  1. Is there evidence of non-linear association between any of the predictors and the response ? To answer this question, for each predictor X, fit a model of the form Y=β0+β1X+β2X2+β3X3+ε.
fit.zn2 <- lm(crim ~ poly(zn, 3))
summary(fit.zn2)
## 
## Call:
## lm(formula = crim ~ poly(zn, 3))
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -4.821 -4.614 -1.294  0.473 84.130 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    3.6135     0.3722   9.709  < 2e-16 ***
## poly(zn, 3)1 -38.7498     8.3722  -4.628  4.7e-06 ***
## poly(zn, 3)2  23.9398     8.3722   2.859  0.00442 ** 
## poly(zn, 3)3 -10.0719     8.3722  -1.203  0.22954    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.372 on 502 degrees of freedom
## Multiple R-squared:  0.05824,    Adjusted R-squared:  0.05261 
## F-statistic: 10.35 on 3 and 502 DF,  p-value: 1.281e-06
fit.indus2 <- lm(crim ~ poly(indus, 3))
summary(fit.indus2)
## 
## Call:
## lm(formula = crim ~ poly(indus, 3))
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -8.278 -2.514  0.054  0.764 79.713 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)        3.614      0.330  10.950  < 2e-16 ***
## poly(indus, 3)1   78.591      7.423  10.587  < 2e-16 ***
## poly(indus, 3)2  -24.395      7.423  -3.286  0.00109 ** 
## poly(indus, 3)3  -54.130      7.423  -7.292  1.2e-12 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.423 on 502 degrees of freedom
## Multiple R-squared:  0.2597, Adjusted R-squared:  0.2552 
## F-statistic: 58.69 on 3 and 502 DF,  p-value: < 2.2e-16
fit.nox2 <- lm(crim ~ poly(nox, 3))
summary(fit.nox2)
## 
## Call:
## lm(formula = crim ~ poly(nox, 3))
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -9.110 -2.068 -0.255  0.739 78.302 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     3.6135     0.3216  11.237  < 2e-16 ***
## poly(nox, 3)1  81.3720     7.2336  11.249  < 2e-16 ***
## poly(nox, 3)2 -28.8286     7.2336  -3.985 7.74e-05 ***
## poly(nox, 3)3 -60.3619     7.2336  -8.345 6.96e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.234 on 502 degrees of freedom
## Multiple R-squared:  0.297,  Adjusted R-squared:  0.2928 
## F-statistic: 70.69 on 3 and 502 DF,  p-value: < 2.2e-16
fit.rm2 <- lm(crim ~ poly(rm, 3))
summary(fit.rm2)
## 
## Call:
## lm(formula = crim ~ poly(rm, 3))
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -18.485  -3.468  -2.221  -0.015  87.219 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    3.6135     0.3703   9.758  < 2e-16 ***
## poly(rm, 3)1 -42.3794     8.3297  -5.088 5.13e-07 ***
## poly(rm, 3)2  26.5768     8.3297   3.191  0.00151 ** 
## poly(rm, 3)3  -5.5103     8.3297  -0.662  0.50858    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.33 on 502 degrees of freedom
## Multiple R-squared:  0.06779,    Adjusted R-squared:  0.06222 
## F-statistic: 12.17 on 3 and 502 DF,  p-value: 1.067e-07
fit.age2 <- lm(crim ~ poly(age, 3))
summary(fit.age2)
## 
## Call:
## lm(formula = crim ~ poly(age, 3))
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -9.762 -2.673 -0.516  0.019 82.842 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     3.6135     0.3485  10.368  < 2e-16 ***
## poly(age, 3)1  68.1820     7.8397   8.697  < 2e-16 ***
## poly(age, 3)2  37.4845     7.8397   4.781 2.29e-06 ***
## poly(age, 3)3  21.3532     7.8397   2.724  0.00668 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.84 on 502 degrees of freedom
## Multiple R-squared:  0.1742, Adjusted R-squared:  0.1693 
## F-statistic: 35.31 on 3 and 502 DF,  p-value: < 2.2e-16
fit.dis2 <- lm(crim ~ poly(dis, 3))
summary(fit.dis2)
## 
## Call:
## lm(formula = crim ~ poly(dis, 3))
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -10.757  -2.588   0.031   1.267  76.378 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     3.6135     0.3259  11.087  < 2e-16 ***
## poly(dis, 3)1 -73.3886     7.3315 -10.010  < 2e-16 ***
## poly(dis, 3)2  56.3730     7.3315   7.689 7.87e-14 ***
## poly(dis, 3)3 -42.6219     7.3315  -5.814 1.09e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.331 on 502 degrees of freedom
## Multiple R-squared:  0.2778, Adjusted R-squared:  0.2735 
## F-statistic: 64.37 on 3 and 502 DF,  p-value: < 2.2e-16
fit.rad2 <- lm(crim ~ poly(rad, 3))
summary(fit.rad2)
## 
## Call:
## lm(formula = crim ~ poly(rad, 3))
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -10.381  -0.412  -0.269   0.179  76.217 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     3.6135     0.2971  12.164  < 2e-16 ***
## poly(rad, 3)1 120.9074     6.6824  18.093  < 2e-16 ***
## poly(rad, 3)2  17.4923     6.6824   2.618  0.00912 ** 
## poly(rad, 3)3   4.6985     6.6824   0.703  0.48231    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.682 on 502 degrees of freedom
## Multiple R-squared:    0.4,  Adjusted R-squared:  0.3965 
## F-statistic: 111.6 on 3 and 502 DF,  p-value: < 2.2e-16
fit.tax2 <- lm(crim ~ poly(tax, 3))
summary(fit.tax2)
## 
## Call:
## lm(formula = crim ~ poly(tax, 3))
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -13.273  -1.389   0.046   0.536  76.950 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)     3.6135     0.3047  11.860  < 2e-16 ***
## poly(tax, 3)1 112.6458     6.8537  16.436  < 2e-16 ***
## poly(tax, 3)2  32.0873     6.8537   4.682 3.67e-06 ***
## poly(tax, 3)3  -7.9968     6.8537  -1.167    0.244    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.854 on 502 degrees of freedom
## Multiple R-squared:  0.3689, Adjusted R-squared:  0.3651 
## F-statistic:  97.8 on 3 and 502 DF,  p-value: < 2.2e-16
fit.ptratio2 <- lm(crim ~ poly(ptratio, 3))
summary(fit.ptratio2)
## 
## Call:
## lm(formula = crim ~ poly(ptratio, 3))
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
## -6.833 -4.146 -1.655  1.408 82.697 
## 
## Coefficients:
##                   Estimate Std. Error t value Pr(>|t|)    
## (Intercept)          3.614      0.361  10.008  < 2e-16 ***
## poly(ptratio, 3)1   56.045      8.122   6.901 1.57e-11 ***
## poly(ptratio, 3)2   24.775      8.122   3.050  0.00241 ** 
## poly(ptratio, 3)3  -22.280      8.122  -2.743  0.00630 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.122 on 502 degrees of freedom
## Multiple R-squared:  0.1138, Adjusted R-squared:  0.1085 
## F-statistic: 21.48 on 3 and 502 DF,  p-value: 4.171e-13
fit.black2 <- lm(crim ~ poly(black, 3))
summary(fit.black2)
## 
## Call:
## lm(formula = crim ~ poly(black, 3))
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -13.096  -2.343  -2.128  -1.439  86.790 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       3.6135     0.3536  10.218   <2e-16 ***
## poly(black, 3)1 -74.4312     7.9546  -9.357   <2e-16 ***
## poly(black, 3)2   5.9264     7.9546   0.745    0.457    
## poly(black, 3)3  -4.8346     7.9546  -0.608    0.544    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.955 on 502 degrees of freedom
## Multiple R-squared:  0.1498, Adjusted R-squared:  0.1448 
## F-statistic: 29.49 on 3 and 502 DF,  p-value: < 2.2e-16
fit.lstat2 <- lm(crim ~ poly(lstat, 3))
summary(fit.lstat2)
## 
## Call:
## lm(formula = crim ~ poly(lstat, 3))
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -15.234  -2.151  -0.486   0.066  83.353 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       3.6135     0.3392  10.654   <2e-16 ***
## poly(lstat, 3)1  88.0697     7.6294  11.543   <2e-16 ***
## poly(lstat, 3)2  15.8882     7.6294   2.082   0.0378 *  
## poly(lstat, 3)3 -11.5740     7.6294  -1.517   0.1299    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 7.629 on 502 degrees of freedom
## Multiple R-squared:  0.2179, Adjusted R-squared:  0.2133 
## F-statistic: 46.63 on 3 and 502 DF,  p-value: < 2.2e-16
fit.medv2 <- lm(crim ~ poly(medv, 3))
summary(fit.medv2)
## 
## Call:
## lm(formula = crim ~ poly(medv, 3))
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -24.427  -1.976  -0.437   0.439  73.655 
## 
## Coefficients:
##                Estimate Std. Error t value Pr(>|t|)    
## (Intercept)       3.614      0.292  12.374  < 2e-16 ***
## poly(medv, 3)1  -75.058      6.569 -11.426  < 2e-16 ***
## poly(medv, 3)2   88.086      6.569  13.409  < 2e-16 ***
## poly(medv, 3)3  -48.033      6.569  -7.312 1.05e-12 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 6.569 on 502 degrees of freedom
## Multiple R-squared:  0.4202, Adjusted R-squared:  0.4167 
## F-statistic: 121.3 on 3 and 502 DF,  p-value: < 2.2e-16

Based on the p-value, the cubic fit is good for predictors:“indus”, “nox”, “age”, “dis”, “ptratio” and “medv”